Ask question

Ask question

All categories

3 years ago

7

Which best describes the wavelength of a simple wave?

1 answer:

Hatshy [7]3 years ago

7 0

Answer:

I think it is A

Explanation: Wavelength is the distance from one crest to another, or from one trough to another, of a wave (which may be an electromagnetic wave, a sound wave, or any other wave). Crest is the highest point of the wave whereas the trough is the lowest.

You might be interested in

Does anyone know? Please help! Thank you!

Vedmedyk [2.9K]

B density will increase

7 0

3 years ago

Read 2 more answers

A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi

sleet_krkn [62]

Answer:

Equilibrium temperature will be $T=52.2684^{\circ}C$

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So $2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)$

$0.334T-3.67=1688-32.031T$

$T=52.2684^{\circ}C$

5 0

3 years ago

What does the law of conservation of energy say

HACTEHA [7]

Answer: states that energy can neither be created nor destroyed - only converted from one form of energy to another.

Explanation:

6 0

3 years ago

Read 2 more answers

A balloon has a volume of 3.0 l at 25Â°c. what is the approximate volume of the balloon at 50Â°c?

mart [117]

Sorry i dont know this one

6 0

3 years ago

A 240 N sphere 0.20 m in radius rolls, without slipping 6.0 m downa

Shalnov [3]

Answer:

The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.

Explanation:

It is given that,

Weight of the sphere, W = 240 N

Radius of the sphere, r = 0.2 m

Angle with the horizontal, $\theta=28^{\circ}$

We need to find the angular speed of the sphere at the bottom of the hill if it starts from rest.

As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.

Gravitational energy = translational energy + rotational energy

So,

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

I is the moment of inertia of the sphere, $I=\dfrac{2}{5}mr^2$

Also, $v=r\omega$

h is the height of the ramp, $h=l\ sin\theta$

$mgl\ sin\theta=\dfrac{1}{2}m(r\omega)^2+\dfrac{1}{2}I\omega^2$

On solving the above equation we get :

$\omega=\sqrt{\dfrac{10gl\ sin\theta}{7r^2}}$

$\omega=\sqrt{\dfrac{10\times 9.8\times 6\ sin(28)}{7(0.2)^2}}$

$\omega=31.39\ rad/s$

So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.

6 0

3 years ago